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(B) Given vectors are $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate the magnitudes: $|

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$\frac{\sqrt{17}}{9}$

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(B) Given vectors are $\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\vec{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$.First,calculate the magnitudes: $|\vec{AB}| = \sqrt{2^2 + 10^2 + 11^2} = \sqrt{4 + 100 + 121} = \sqrt{225} = 15$.$|\vec{AD}| = \sqrt{(-1)^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.Let $\alpha$ be the angle between $\vec{AB}$ and $\vec{AD}$. Then $\cos \alpha = \frac{\vec{AB} \cdot \vec{AD}}{|\vec{AB}| |\vec{AD}|} = \frac{(2)(-1) + (10)(2) + (11)(2)}{15 	imes 3} = \frac{-2 + 20 + 22}{45} = \frac{40}{45} = \frac{8}{9}$.Since $\vec{AD'}$ is obtained by rotating $\vec{AD}$ by an angle $	heta$ in the plane of the parallelogram such that $\vec{AD'} \perp \vec{AB}$,the angle between $\vec{AD'}$ and $\vec{AB}$ is $90^\circ$.The angle between $\vec{AD}$ and $\vec{AD'}$ is $	heta$. Thus,$	heta = |\alpha - 90^\circ|$ or $	heta = |90^\circ - \alpha|$.Therefore,$\cos 	heta = \cos(90^\circ - \alpha) = \sin \alpha$.Since $\cos \alpha = 8/9$,we have $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (8/9)^2} = \sqrt{1 - 64/81} = \sqrt{17/81} = \frac{\sqrt{17}}{9}$.

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